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\title{复变函数练习5.1-5.2 - 解析函数的洛朗展式、孤立奇点 }
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\date{2024 年 5 月 27 日}
%\date{March 9, 2021}

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\begin{document}

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\begin{enumerate}

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\item  %Problem 01
（洛朗定理）圆环 $H:r<|z-a|<R$ 内的解析函数 $f(z)$ 可以唯一地展开成双边幂级数
$$f(z)=\sum\limits_{n=-\infty}^{\infty} c_n(z-a)^n,$$
其中系数为沿着圆周 $\Gamma: |\zeta-a|=\rho \, (r<\rho<R)$ 的复积分，
$$c_n = \frac{1}{2\pi i} \int_\Gamma \frac{f(\zeta)}{(\zeta-a)^{n+1}}d\zeta,\,\, n=0,\pm 1, \pm 2, \cdots.  $$


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\item  %Problem 02
（例子5.1）求函数 $f(z)=\frac{1}{(z-1)(z-2)}$ 在下述三个区域内的洛朗展式，
\begin{enumerate}
\item  圆 $|z|<1$; 
\item  圆环 $1<|z|<2$; 
\item  圆环 $2<|z|<+\infty$. 
\end{enumerate}

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\item  %Problem 03
（施瓦茨引理）设 $f(z)$ 是从单位圆映到自身的解析函数，设 $f(0)=0$.  则
\begin{enumerate}
\item  在单位圆内恒有 $|f(z)|\le |z|$, 且有 $|f'(0)|\le 1$. 
\item  若 $|f'(0)|=1$, 或存在 $0<|z_0|<1$ 使得 $|f(z_0)|=|z_0|$, 则存在实常数 $\alpha$ 使得 $f(z)=e^{i\alpha} z$.  
\end{enumerate}

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\item  %Problem 04
找出下述函数的奇点，并区分为可去奇点、极点和本质奇点。
\begin{enumerate}
\item  $f(z)=\frac{e^z-1}{z}$. 
\item  $f(z)=\frac{5z+1}{(z-1)(2z+1)^2}$. 
\item  $f(z)=\frac{1}{\sin z}$. 
\item  $f(z)=e^{\frac{1}{z}}$. 
\item  $f(z)=\sin\frac{1}{z}$. 
\end{enumerate}

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\item  %Problem 05
（魏尔斯特拉斯定理）设 $a$ 为函数 $f(z)$ 的本质奇点，则对于任何复数或无穷大 $A$, 存在收敛于 $a$ 的数列 $z_n$ 使得 $\lim\limits_{n\to\infty} f(z_n)=A$. 

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\item  %Problem 06
（皮卡大定理）设 $a$ 为函数 $f(z)$ 的本质奇点，则对于任何复数 $A$, 除掉可能一个值 $A_0$ 之外，
存在收敛于 $a$ 的数列 $\{z_n\}$ 使得 $f(z_n)=A\,(n=1,2,\cdots)$. 

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\item  %Problem 07
（例子5.12）以函数 $f(z)=e^{\frac{1}{z}}$ 为例，说明皮卡大定理。

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\end{enumerate}


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\end{document}

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